Root decomposition implies semisimple

Question

I am trying to solve exercise 6.4 from "Kirillov: An Introduction to Lie Groups and Lie Algebras". The exercise states:

Let $\mathfrak{g}$ be a complex Lie algebra which has a root decomposition \begin{align} \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{\alpha \in R} \mathfrak{g}_\alpha \end{align} where $R$ is a finite subset in $\mathfrak{h}^* \setminus \{ 0 \}$, $\mathfrak{h}$ is commutative and for $h \in \mathfrak{h}$, $x \in \mathfrak{g}_\alpha$, we have $[h,x]=\langle h,\alpha \rangle x$. How that then $\mathfrak{g}$ is semisimple, and $\mathfrak{h}$ is a Cartan subalgebra.

$\mathfrak{h}^*$ is the ordinary dual vector space of the vector space $\mathfrak{h}$. $\langle \bullet, \bullet \rangle$ denotes the pairing $\mathfrak{h} \times \mathfrak{h}^* \rightarrow \mathbb{C}$.

Once it is shown that $\mathfrak{g}$ is semisimple, it should follow easily that $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$. However I cannot prove the semisimplicity.

My observations/assumptions so far: One must assume $R \neq \emptyset$, otherwise $\mathfrak{g}$ may be not semisimple. "Root decomposition" does not imply properties apart from the ones mentioned in the end of the sentence. One might hope to show that the Killing form $K$ on $\mathfrak{g}$ is non-degenerate. Clearly $K(h,h) \neq 0$ for all $h \in \mathfrak{h}$. But why is $K$ non-degenerate on the $\mathfrak{g}_\alpha$?

(This is a clarification of question If $\mathfrak{g}$ admits a decomposition then it is semisimple )

Answer 1

I don't think that this is true in the form you state it, i.e. that additional assumptions on the decomposition are needed. As far as I can see, the following are counter examples to the current claim: Take $\mathfrak g:=\mathfrak{gl}(n,\mathbb C)$ with $\mathfrak h$ the subspace of diagonal matrices and the usual root spaces $\mathfrak g_{\alpha}$ (as for $\mathfrak{sl}(n,\mathbb C)$). Then this satisfies all conditions but $\mathfrak g$ is reductive rather than semi-simple. (This example also shows that you cannot deduce that $K(h,h)\neq 0$.) More drastically, you can take the Lie algebra $\mathfrak b\subset\mathfrak{gl}(n,\mathbb C)$ of upper triangular matrices with the induced decomposition. So you leave $\mathfrak h$ as it is and only keep positive root spaces. Again, this satisfies all conditions, but $\mathfrak b$ is solvable.

Sufficient additional conditions should for example be:

1. $R$ spans $\mathfrak h^*$, for $\alpha\in R$ also $-\alpha\in R$ and for $X\in\mathfrak g_{\alpha}$, there is $Y\in\mathfrak g_{-\alpha}$ such that $[X,Y]\neq 0$.

2. $\mathfrak h\subset [\mathfrak g,\mathfrak g]$

Assuming these or similar conditions, there is a direct argument to prove the claim. The main issue here is that the given decomposition is the joint eigenvalue decomposition for the adjoint actions of the elements of $\mathfrak h$. From the fact that the projection onto an eigenspace of an operator can be written as a polynomial in the operator, one deduces that if an ideal $I$ contains an element $H+\sum_{\alpha}X_\alpha$, then each of the summands lies in $I$. From there, you show step by step that $I$ contains all of $\mathfrak g$.